Binomial coefficient equivalence

Question

Can someone explain to me why these 2 formulas are equivalent: $${n \choose k} = {n \choose n-k}$$

Answer 1

Think of it this way: Say we have a collection of $n$ things and will be taking $k$ of them with us. Then we can either choose which $k$ to take with us--which we can do in $\binom{n}{k}$ ways--or choose which $n-k$ not to take with us--which we can do in $\binom{n}{n-k}$ ways. Both approaches yield exactly the same result, so the counts are the same.

Answer 2

If you are choosing $k$ things out of a whole of $n$ is the same as not choosing the remaining $n-k$.

Answer 3

Consider a set $A$, and the cardinality of $A$, $|A|=n,\ n\in\mathbb{Z}^{\geqslant}$. The two subsets, $\mathcal{P}_r(A)$ and $\mathcal{P}_{n-r}(A)$ of the power set of $A$, $\mathcal{P}(A)$, have the same cardinality, i.e., $$ {n \choose r}={n \choose n-r}. $$

Answer 4

You can see they are the same after a couple arithmetic operations.

${n \choose k} = {n! \over k!(n-k)!}$

${n \choose n-k} = {n! \over (n-k)!(n-(n-k))!} = {n! \over (n-k)!(n-n+k))!} = {n! \over (n-k)!(k)!} = {n! \over k!(n-k)!}$